What Causes Electric Field Breakdown in P-N Diodes?

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SUMMARY

The discussion focuses on the mechanisms of electric field breakdown in P-N diodes, particularly under reverse bias conditions. When the breakdown voltage is reached, the electric field from the depletion zone causes covalent bonds to break, allowing current to flow due to the generation of electron-hole pairs. Key processes involved in breakdown include impact ionization and tunneling, as seen in Zener diodes. Additionally, the discussion touches on the behavior of rectified AC signals and the importance of oscilloscope settings in visualizing these signals accurately.

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  • P-N junction theory
  • Understanding of covalent bonding in semiconductors
  • Knowledge of rectification processes (half-wave and full-wave)
  • Familiarity with oscilloscope operation and settings
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Electrical engineers, semiconductor physicists, students studying electronics, and anyone involved in the design or analysis of P-N diodes and rectification circuits.

quietrain
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i am not sure if this is considered electrodynamics under classical forum but oh well, here goes

i was reading wiki about p-n diodes and its that for reverse bias diode, because of the anode attracting the holes WITH the cathode attracting the electrons, the depletion zone widens and causes an increase in voltage barrier, so current will be harder to flow.

so when breakdown voltage is reached, the electric field due to the depletion zone breaks down and current flows. so what exactly does the electric field breakdown means?

also regarding rectified ac, i realize that when i rectify an AC source half or full wave, it seems that the oscilloscope shows

ac.png


with the blue part rectified wave, and red the ac transformer source.

so why issnt the blue part directly fitting the red curve? i have to adjust the position of the blue curve upwards on the oscilloscope to fit nicely.

also when full wave rectifying, what gives the extra rectified upward curve? is it the diode causing the original downward ac curve to flip up?

thanks loads sry for the long question
 
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in reverse biasing when we apply high reverse voltage then ,electron in the diode bonded by covalent bond gets free due to the brockage of covalent bond & breakdown ocures.the covalent bond are broke up due to the high electric field across the junction,the break down voltage depend on the doping of the semiconducter...
 
nikhil khatri said:
in reverse biasing when we apply high reverse voltage then ,electron in the diode bonded by covalent bond gets free due to the brockage of covalent bond & breakdown ocures.the covalent bond are broke up due to the high electric field across the junction,the break down voltage depend on the doping of the semiconducter...

why does having the covalent bond breaking up allows for electricity to pass through?

does covalent bonding has anything to do with the holes and electrons in the p-n semiconductor diode?
 
quietrain said:
why does having the covalent bond breaking up allows for electricity to pass through?

does covalent bonding has anything to do with the holes and electrons in the p-n semiconductor diode?

yes ,when covalent bond breaks then electron leave its place from the bond & thus creat a hole there in this manner when one bond break up then a pair of hole& electron is generated.
 
quietrain said:
so when breakdown voltage is reached, the electric field due to the depletion zone breaks down and current flows. so what exactly does the electric field breakdown means?

Breakdown can occur through various processes, but generally carriers are generated by impact ionization (avalanche breakdown), or because they can tunnel through the junction (like in a Zener diode). You can check these terms on wikipedia or google for more details.

so why issnt the blue part directly fitting the red curve? i have to adjust the position of the blue curve upwards on the oscilloscope to fit nicely.

Is your oscilloscope in AC mode by any chance? In AC mode, the oscilloscope will remove the DC component, so the rectified wave is "centered". If you set it to DC mode, the curves should match up.

also when full wave rectifying, what gives the extra rectified upward curve? is it the diode causing the original downward ac curve to flip up?
Yes, sort of. In the half-wave rectifier, the diode stops the current one way. In the full-wave rectifier, there's always a diode that let's the current through.
 
nikhil khatri said:
yes ,when covalent bond breaks then electron leave its place from the bond & thus creat a hole there in this manner when one bond break up then a pair of hole& electron is generated.

so why does the p-n diode already has a high concentrations of holes at one side and another set of high concentration of electrons at another side as explained in wikipedia?

is it due to doping?
 
caffenta said:
Breakdown can occur through various processes, but generally carriers are generated by impact ionization (avalanche breakdown), or because they can tunnel through the junction (like in a Zener diode). You can check these terms on wikipedia or google for more details.



Is your oscilloscope in AC mode by any chance? In AC mode, the oscilloscope will remove the DC component, so the rectified wave is "centered". If you set it to DC mode, the curves should match up.


Yes, sort of. In the half-wave rectifier, the diode stops the current one way. In the full-wave rectifier, there's always a diode that let's the current through.

ah isee thanks!
 
quietrain said:
so why does the p-n diode already has a high concentrations of holes at one side and another set of high concentration of electrons at another side as explained in wikipedia?

is it due to doping?

yes it depend on doping that which region will contain holes & which will contain electron ...
 
nikhil khatri said:
yes it depend on doping that which region will contain holes & which will contain electron ...

ah i see thank you!
 

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